🎉 #1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence

Author: GoogTech

Diff for: #1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence/Solution.java

@@ -0,0 +1,27 @@
+/*
+ * @Author: Goog Tech
+ * @Date: 2020-08-29 17:48:14
+ * @LastEditTime: 2020-08-29 17:48:33
+ * @Description: https://leetcode-cn.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/
+ * @FilePath: \leetcode-googtech\#1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence\Solution.java
+ * @WebSite: https://algorithm.show/
+ */
+
+class Solution {
+    public int isPrefixOfWord(String sentence, String searchWord) {
+        // 切割字符串并将其转化为字符数组
+        String[] strs = sentence.split(" ");
+        // 获取检索词的长度
+        int length = searchWord.length();
+        // 遍历字符数组
+        for(int i = 0; i < strs.length; i++) {
+            // 判断当前字符元素的长度是否大于检索词的长度,并且切割后的字符元素是否与检索词相同
+            if(strs[i].length() >= length && strs[i].substring(0, length).equals(searchWord)) {
+                // 若相同则返回句子 sentence 中该单词所对应的下标(下标从 1 开始)
+                return i + 1;
+            }
+        }
+        // 无果
+        return -1;
+    }
+}

Diff for: #1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence/Solution.py

@@ -0,0 +1,26 @@
+'''
+Author: Goog Tech
+Date: 2020-08-29 17:48:18
+LastEditTime: 2020-08-29 17:48:41
+Description: https://leetcode-cn.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/
+FilePath: \leetcode-googtech\#1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence\Solution.py
+WebSite: https://algorithm.show/
+'''
+
+class Solution(object):
+    def isPrefixOfWord(self, sentence, searchWord):
+        """
+        :type sentence: str
+        :type searchWord: str
+        :rtype: int
+        """
+        # 切割字符串并将其转化为字符数组
+        words = sentence.split(' ')
+        # 遍历字符数组
+        for i in range(len(words)):
+            # 判断当前字符元素的长度是否大于检索词的长度,并且切割后的字符元素是否与检索词相同
+            if len(words[i]) >= len(searchWord) and words[i][0 : len(searchWord)] == searchWord:
+                # 若相同则返回句子 sentence 中该单词所对应的下标(下标从 1 开始)
+                return i + 1
+        # 无果
+        return -1